Given dy/dx=35x^6-9*e^3x find y.

To find out the original, when knowing it's derivative,
we'll have to integrate the expression of derivative.

We'll
determine the indefinite integral of f'(x)=35x^6-9*e^3x

Int
dy = y + C

Int
(35x^6-9*e^3x)dx

We'll use the property of the indefinite
integral, to be additive:

Int (35x^6-9*e^3x)dx  =Int
(35x^6)dx - Int (9e^3x)dx

Int (35x^6)dx = 35*x^(6+1)/(6+1)
+ C

Int (35x^6)dx = 35x^7/7 +
C

Int (35x^6)dx = 5x^7 + C
(1)

Int 9e^3x dx = 9*e^3x/3 +
C

Int 9e^3x dx = 3e^3x + C
(2)

We'll subtract (2) from (1) and we'll
get:

Int (35x^6-9*e^3x)dx = 5x^7 - 3e^3x  +
C

So, the primitive function is: y = 5x^7 -
3e^3x + C